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Calculus_I_Chapter_1 (Printed version).pdf

  1. 1. CALCULUS I - AY2022/23 Chapter 1: Functions, Limits, and Continuity CALCULUS I - AY2022/23 1 / 109
  2. 2. Outline 1 Functions 2 Limits 3 Continuity CALCULUS I - AY2022/23 2 / 109
  3. 3. Summary Functions are fundamental to the study of calculus. In this section we study the following topics: Functions and their graphs Combining functions Important types of functions CALCULUS I - AY2022/23 3 / 109
  4. 4. • What is a function? Definition. A function f from a set D to a set Y is a rule that assigns to each element x in D a unique (single) element, called f(x), in Y. The arrow diagram for f looks as follows. CALCULUS I - AY2022/23 4 / 109
  5. 5. • A function as a kind of machine The set D of all possible input values is called the domain of the function, while set of all possible values of f(x) as x varies throughout D is called the range of the function. So we can imagine a function f as a machine which produces an output value f(x) in its range whenever we feed it an input x from its domain. CALCULUS I - AY2022/23 5 / 109
  6. 6. • How to represent functions? We will mainly study functions for which the sets D and Y are sets of real numbers. There are many different ways of describing functions For example, by a description in words (verbally) by a table of values (numerically) by a graph (visually) by an explicit formula (algebraically) CALCULUS I - AY2022/23 6 / 109
  7. 7. • Visualizing a function The most common method for visualizing a function is its graph. The graph of a function For a function f with domain D, its graph is the set of all points (x, y) in the xy-plane, with y = f(x) and x is in the domain of f. We write G := {(x, f(x)) : x ∈ D}. CALCULUS I - AY2022/23 7 / 109
  8. 8. • Examples Domain convention If a function is given by a formula and the domain is not stated explicitly, then the convention is that the domain is the set of all numbers for which the format makes sense and defines a real number. Exercise What are the domains of the following functions? (a) f(x) = 1 1 − cos2 x (b) g(x) = 1 √ x − 1 − √ 1 − x Answer. (a) x 6= kπ (k = 0, ±1, ±2, ...); (b) x = ∅. CALCULUS I - AY2022/23 8 / 109
  9. 9. • Piecewise-defined functions The functions defined by different formulas in different parts of their domains are called piecewise-defined functions. Example (Absolute value function) |x| =    x, x ≥ 0 −x, x < 0. The graph is as follows. CALCULUS I - AY2022/23 9 / 109
  10. 10. • Even and odd functions: Symmetry Definition. A function y = f(x) is called even, if f(−x) = f(x), odd, if f(−x) = −f(x), for every x in its domain. CALCULUS I - AY2022/23 10 / 109
  11. 11. Note. The graph of an even function is symmetric about the y-axis, while the graph of an odd function is symmetric about the origin. Exercise Determine whether each of the following functions is even, odd, or neither even nor odd. (a) f(x) = x5 + x (b) g(x) = 1 − x4 (c) h(x) = 2x − x2 . Answer. (a) odd; (b) even; (c) neither even nor odd. CALCULUS I - AY2022/23 11 / 109
  12. 12. • Increasing and decreasing functions Definition. A function y = f(x) defined on an interval I is called increasing, if f(x1) < f(x2), for any points x1 < x2 in I, decreasing, if f(x1) > f(x2), for any points x1 < x2 in I. Note. A function is increasing if its graph climbs or rises as you move from left to right. A function is decreasing if its graph descents or falls as you move from left to right. CALCULUS I - AY2022/23 12 / 109
  13. 13. • Combining functions Two functions can be composed at a point x if the value of one function at x lies in the domain of the other. More precisely, suppose that we have functions g : X → Y, f : Y → Z, and the range of g is a subset of the domain of f. Definition. The composite function f ◦g (reading as “f composed with g”) is defined by (f ◦ g)(x) = f g(x) , for x from the domain of g CALCULUS I - AY2022/23 13 / 109
  14. 14. The arrow diagram for f ◦ g looks as follows. CALCULUS I - AY2022/23 14 / 109
  15. 15. Example Let f(x) = x2 + 3 and g(x) = √ x. Determine the function f ◦ g and its domain. Solution. We have (f ◦ g)(x) = f(g(x)) = [g(x)]2 + 3 = ( √ x)2 + 3 = x + 3. Its domain is D = {x ∈ R : x 6= 0}. Question: What is g ◦ f and its domain? Answer. (g ◦ f)(x) = √ x2 + 3 and D = (−∞, ∞). CALCULUS I - AY2022/23 15 / 109
  16. 16. Note. It is possible to take the composition of three or more functions. For instance, the composite function f ◦ g ◦ h is found by first applying h, then g, and then f as follows: (f ◦ g ◦ h)(x) = f g(h(x)) . CALCULUS I - AY2022/23 16 / 109
  17. 17. • Important types of functions Polynomials A polynomials is a function of the form P(x) = anxn + an−1xn−1 + · · · + a2x2 + a1x + a0, where n is a non-negative integer and the numbers a0, a1, a2, · · · , an are real constants (called the coefficients of the polynomial). The domain of any polynomial is R = (−∞, ∞). If the leading coefficient an 6= 0, then n is called the degree of the polynomial. CALCULUS I - AY2022/23 17 / 109
  18. 18. - A polynomial of degree 1 is of the form P(x) = mx + b and it is called a linear function (m is a slope). - A polynomial of degree 2 is of the form P(x) = ax2 + bx + c and it is called a quadratic function. - A polynomial of degree 3 is of the form P(x) = ax3 + bx2 + cx + d and it is called a cubic function. CALCULUS I - AY2022/23 18 / 109
  19. 19. Power functions A function f(x) = xa, where a is a constant, is called a power function. There are several important cases to consider. - a = n, a positive number CALCULUS I - AY2022/23 19 / 109
  20. 20. - a = −1 or a = −2 - a = 1 2 , 1 3 . CALCULUS I - AY2022/23 20 / 109
  21. 21. Rational functions A rational function is a quotient or ratio f(x) = P(x) Q(x) , where P and Q are polynomials. The domain of a rational functions is the set {x : Q(x) 6= 0}. CALCULUS I - AY2022/23 21 / 109
  22. 22. Algebraic functions A function is called an algebraic function if it can be constructed from polynomials using algebraic operations (such as addition, subtraction, multiplication, division, and taking roots). Figures below display the graphs of some algebraic functions. CALCULUS I - AY2022/23 22 / 109
  23. 23. Trigonometric functions Recall that a function f(x) is said to be periodic if there is a positive number T such that f(x + T) = f(x), for every value of x. The smallest such value of T is called the period of f. We have six basic trigonometric functions (cosine, sine, tangent, cotangent, secant, and cosecant): cos x, sin x, tan x, cot x, sec x = 1 cos x , csc x = 1 sin x . The graphs of these functions are showed below, where the shading for each function indicates its periodicity. CALCULUS I - AY2022/23 23 / 109
  24. 24. - Functions cos x and sin x. CALCULUS I - AY2022/23 24 / 109
  25. 25. - Functions tan x and sec x. CALCULUS I - AY2022/23 25 / 109
  26. 26. - Functions csc x and cot x. CALCULUS I - AY2022/23 26 / 109
  27. 27. Exponential functions The exponential functions are the functions of the form f(x) = ax , where the base a 0 and a 6= 1, is a constant. All exponential functions have domain (−∞, ∞) and range (0, ∞). So these functions never assume the value 0. CALCULUS I - AY2022/23 27 / 109
  28. 28. Logarithmic functions The logarithmic functions are the functions of the form f(x) = logax, where the base a 0 and a 6= 1, is a constant. They are the inverse functions of the exponential functions. The domain is (0, ∞) and the range is (−∞, ∞). CALCULUS I - AY2022/23 28 / 109
  29. 29. Outline 1 Functions 2 Limits 3 Continuity CALCULUS I - AY2022/23 29 / 109
  30. 30. Summary In this section, we study the following topics: Tangents to curves Limits, one-sided limits Limit laws Infinite limits The concept of a limit is a central idea that distinguishes calculus from algebra and trigonometry. It is fundamental to finding the tangent to a curve. CALCULUS I - AY2022/23 30 / 109
  31. 31. A tangent to a curve A tangent to a curve is a line that touches the curve. For a special case, when a curve is a circle, a line is called tangent to a circle if it meets the circle at precisely one point. CALCULUS I - AY2022/23 31 / 109
  32. 32. • Limit and the tangent to a curve To define tangency for general curves, we use an approach that analyzes the behavior of the secant lines that pass through P and nearby points Q as Q moves toward P along the curve. CALCULUS I - AY2022/23 32 / 109
  33. 33. • Limits In order to solve tangent problems we must be able to find limits. Limit is one of the most fundamental concepts in calculus (and, in fact, is one of the most important ideas in all of mathematics). What is a limit? Let’s think about the function f(x) = sin x x , and focus on how it looks close to the point x = 0. CALCULUS I - AY2022/23 33 / 109
  34. 34. Question. What is the domain of f(x)? Answer. R {0}. Question. What is the value of f(0)? Answer. The point 0 does not lie in the domain of f, so f does not assign any value to the point x = 0. The key idea Even though f(x) = sin x x has no value defined at the point x = 0, we can still look at the values f takes at points close to x = 0, and see if those values are approaching some fixed value as x gets closer and closer to 0. That is precisely what the limit of a function f(x) at a point a is: a number L which the values of the function get “closer and closer” to as the variable x gets “closer and closer” to a. CALCULUS I - AY2022/23 34 / 109
  35. 35. Let’s calculate some values to see what happens as x gets close to 0: Approaching 0 from the left: x -0.25 -0.1 -0.05 -0.01 -0.005 -0.001 sin x x 0.9896 0.9983 0.9996 0.999983 0.999996 0.999998 Approaching 0 from the right: x 0.25 0.1 0.05 0.01 0.005 0.001 sin x x 0.9896 0.9983 0.9996 0.999983 0.999996 0.999998 Comment. We can only guess sin x x is closer and closer to 1 as x is closer and closer to 0? CALCULUS I - AY2022/23 35 / 109
  36. 36. The graph of function f(x) = sin x x is as follows. CALCULUS I - AY2022/23 36 / 109
  37. 37. • Definition of a limit Definition. Let f(x) be a function defined on some open interval that contains a, except possibly at a itself. We say that the limit of f(x) as x tends to a is L, and write lim x→a f(x) = L, if for every number ε 0, there exists a corresponding number δ 0, such that 0 |x − a| δ =⇒ |f(x) − L| ε. CALCULUS I - AY2022/23 37 / 109
  38. 38. • We can have the following figure. • By the graph interpretation, it looks as follows. CALCULUS I - AY2022/23 38 / 109
  39. 39. • A typical procedure to prove lim x→a = L by definition To prove a particular limit our task is to explain why we can always choose an appropriate δ 0, no matter what ε 0 is. (1) Let ε 0 be any number. Choose an appropriate number δ 0. (We need to find δ by several-step calculations - this is the important part of the proof) (Note that δ depends on ε, that is formally speaking, we need to find a good function δ(ε) : R+ → R+. Here R+ = {x ∈ R : x 0}.) (2) Verify (prove) δ works, that is 0 |x − a| δ =⇒ |f(x) − L| ε. Conclude that lim x→a = L. CALCULUS I - AY2022/23 39 / 109
  40. 40. Example Prove that lim x→3 (4x − 5) = 7. Solution. Preliminary analysis of the problem (guessing a value for δ). Let ε 0 be given. We want to find a number δ 0 such that 0 |x − 3| δ =⇒ |(4x − 5) − 7| | {z } 4|x−3| ε. So we need to find δ so that 0 |x − 3| δ =⇒ 4|x − 3| ε ⇐⇒ |x − 3| ε 4 . To make sure that |x − 3| ε 4 , we should choose δ = ε 4 . CALCULUS I - AY2022/23 40 / 109
  41. 41. Proof (showing that this δ works). Given ε 0, choose δ = ε 4 . In this case, if 0 |x − 3| ε 4 , then |(4x − 5) − 7| = 4|x − 3| 4δ = 4 · ε 4 = ε. Thus 0 |x − 3| δ =⇒ |(4x − 5) − 7| ε, which shows, by the definition of a limit, that lim x→3 (4x − 5) = 7. Comment. In the solution there were two stages: guessing proving. - First we made a preliminary analysis that enabled us to guess a value for δ. - Next we had to go back and prove in a careful, logical fashion that we had made a correct guess. CALCULUS I - AY2022/23 41 / 109
  42. 42. • How to show that the limit does not exists? Let us consider the following function f(x). How would we use the ε-δ technique to show that the limit lim x→0 f(x) does not exist? In general, it is not simple. We will come back to this problem later. CALCULUS I - AY2022/23 42 / 109
  43. 43. One-sided limits Example The Heaviside function H is defined by H(x) =    0, t 0 1, t ≥ 0. This function is named after Oliver Heaviside (1850-1925) (who was an English electrical engineer, mathematician, and physicist) and can be used to describe an electric current that is switched on at time t = 0. Its graph is shown below CALCULUS I - AY2022/23 43 / 109
  44. 44. It is clear that as t tends to 0 from the left, H(t) approaches 0, while as t tends to 0 from the right, H(t) approaches 1. Moreover, there is no single number that H(t) approaches as t tends to 0. Therefore lim t→0 H(t) does not exist. We indicate the situation symbolically by writing lim t→0 t0 H(t) = lim t→0− H(t) = 0, and lim t→0 t0 H(t) = lim t→0+ H(t) = 1. CALCULUS I - AY2022/23 44 / 109
  45. 45. One-sided limits We also have notions of one-sided limits. Definition of one-side limits lim x→a− f(x) = L if for every number ε 0 there is a number δ 0 such that a − δ x a =⇒ |f(x) − L| ε. lim x→a+ f(x) = L if for every number ε 0 there is a number δ 0 such that a x a + δ =⇒ |f(x) − L| ε. CALCULUS I - AY2022/23 45 / 109
  46. 46. The definitions of one-sided limits are illustrated in the following figure. CALCULUS I - AY2022/23 46 / 109
  47. 47. • The trick estimation In general, it is not always easy to prove that limit statements are true using the ε, δ definition. Sometimes we can prove complicated limits using a trick called estimation. • The main idea of estimation Estimation is a useful trick which can sometimes help us prove complicated limits. It basically means replacing a complicated piece of some expression by something larger, but easier to understand. Actually, this trick - replacing complicated functions by simpler, larger functions - is useful in many different parts of mathematics, not just calculus. CALCULUS I - AY2022/23 47 / 109
  48. 48. Example Use estimation to prove that lim x→3 x3 = 27. Solution. Guessing a value for δ. Let ε 0 be given. We want to find a number δ 0 so that 0 |x − 3| δ =⇒ |x3 − 27| ε. To connect |x3 − 27| with |x − 3|, we write |x3 − 27| = |(x − 3)(x2 + 3x + 9)| = |x − 3| · |x2 + 3x + 9|. Then we want 0 |x − 3| δ =⇒ |x − 3| · |x2 + 3x + 9| ε. Question. How can we process from |x − 3| · |x2 + 3x + 9| to |x − 3|? CALCULUS I - AY2022/23 48 / 109
  49. 49. Answer. We can try to replace the complicated term |x2 + 3x + 9| by a constant C 0 so that |x2 + 3x + 9| C around the point x = 3. If we can find such a C, then |x3 − 27| = |x − 3| · |x2 + 3x + 9| C|x − 3|. In this case, we can make C|x − 3| ε by taking |x − 3| ε C , and so we could choose δ = ε C . However, |x2 + 3x + 9| has no upper bound in its domain R. We can only find such a number C if we restrict x to be in some interval centered at 3. CALCULUS I - AY2022/23 49 / 109
  50. 50. In fact, since we are interested only in values of x that are close to 3, it is reasonable to assume that x is within a distance 1 from 3, that is |x − 3| 1. Then we have 2 x 4, which gives 19 x2 + 3x + 9 37. Thus |x2 + 3x + 9| 37, and hence C = 37 is a suitable choice for the constant. But now there are two restrictions on |x − 3|, namely |x − 3| 1 and |x − 3| ε C = ε 37 . To make sure that both of these inequalities are satisfied, we take δ to be the smaller of the two numbers 1 and ε 37 . The notation for this is δ = min n 1, ε 37 o . CALCULUS I - AY2022/23 50 / 109
  51. 51. Showing that this δ works. Given ε 0, let δ = min n 1, ε 37 o . If 0 |x − 3| δ, then |x − 3| 1 =⇒ 2 x 4 =⇒ |x2 + 3x + 9| 37. We also have |x − 3| ε 37 , and so |x3 − 27| = |x − 3| · |x2 + 3x + 9| ε 37 · 37 = ε. Therefore, lim x→3 x3 = 27. CALCULUS I - AY2022/23 51 / 109
  52. 52. Comment. It is not always easy to prove that limit statements are true using the ε, δ definition. Fortunately, this is unnecessary, the limits of complicated functions can be found rigorously from the Limit Laws without resorting to the definition directly. CALCULUS I - AY2022/23 52 / 109
  53. 53. • Limit Laws Theorem Let L, M, a, and k be real numbers and lim x→a f(x) = L, lim x→a g(x) = M. Then the following rules are true. Sum/Difference rules: lim x→a f(x) ± g(x) = L ± M Constant multiple rule: lim x→a k · f(x) = k · L Product rule: lim x→a f(x) · g(x) = L · M Quotient rule: lim x→a f(x) g(x) = L M , provided that M 6= 0 Power rule: lim x→a [f(x)]n = Ln , n is a positive integer Root rule: lim x→a n p f(x) = n √ L, n is a positive integer (If n is even, we assume that f(x) ≥ 0 for all x in an interval containing a.) CALCULUS I - AY2022/23 53 / 109
  54. 54. • The triangle inequality In many cases, for simplifying complicated expressions and proving limits, we also us the following triangle inequality. Triangle Inequality For any numbers a, b, we always have |a + b| ≤ |a| + |b|. The triangle inequality is useful in mathematical analysis for determining the best upper estimate on the sum of two or more numbers. Here we can use the triangle inequality to prove, for example, the Sum rule above. CALCULUS I - AY2022/23 54 / 109
  55. 55. • Direct Substitution Property We can check that if f is a polynomial or a rational function and a is in the domain of f, then lim x→a f(x) = f(a). Note, however, that not all limits can be evaluated by direct substitution. CALCULUS I - AY2022/23 55 / 109
  56. 56. Example Evaluate the limit lim t→0 √ t2 + 9 − 3 t2 . Solution. We cannot apply the Quotient rule immediately, since the limit of the denominator is 0. But we can rationalize the numerator as follows. lim t→0 √ t2 + 9 − 3 t2 = lim t→0 √ t2 + 9 − 3 t2 · √ t2 + 9 + 3 √ t2 + 9 + 3 = lim t→0 (t2 + 9) − 9 t2( √ t2 + 9 + 3) = lim t→0 t2 t2( √ t2 + 9 + 3) = lim t→0 1 √ t2 + 9 + 3 = 1 lim t→0 p t2 + 9 + 3 = 1 3 + 3 = 1 6 . CALCULUS I - AY2022/23 56 / 109
  57. 57. • The one-sided limits theorem Some limits are best calculated by first finding the left- and right-hand limits. The following result gives an alternative method to prove the existence of a (two-sided) limit via one-sided limits exist. Theorem (The one-sided limits) Let f : D → R be a function. Then the limit lim x→a f(x) exists if and only if the two one-sided limits both exist and are equal. In this case lim x→a f(x) = lim x→a− f(x) = lim x→a+ f(x). Note. When computing one-sided limits, we can use the fact that the Limit Laws also hold for this type of limits. CALCULUS I - AY2022/23 57 / 109
  58. 58. • Limit does not exist Now we come back to the question: How to prove that a limit does not exist? The one-sided limits theorem above gives us the following good way to prove lim x→a f(x) does not exit. Theorem (Limit does not exist) If either of the one-side limits lim x→a− f(x), lim x→a+ f(x) does not exits, or both one-side limits exit, but they are not equal, then so does the limit lim x→a f(x). CALCULUS I - AY2022/23 58 / 109
  59. 59. Example Prove that the following limit does not exist. lim x→0 |x| x . Solution. We compute the one-side limits lim x→0+ |x| x = lim x→0+ x x = lim x→0+ 1 = 1 lim x→0− |x| x = lim x→0− −x x = lim x→0− (−1) = −1. CALCULUS I - AY2022/23 59 / 109
  60. 60. Since the right- and left-hand limits exist, but they are different, we conclude that lim x→0 |x| x does not exist. CALCULUS I - AY2022/23 60 / 109
  61. 61. The next two theorems give two additional properties of limits. Theorem Suppose There exist the limits lim x→a f(x) = A lim x→a g(x) = B f(x) ≤ g(x) for all x in some open interval containing a, except possibly at x = a itself. Then lim x→a f(x) ≤ lim x→a g(x), that is A ≤ B. CALCULUS I - AY2022/23 61 / 109
  62. 62. Theorem (The Squeeze Theorem) Suppose f(x) ≤ g(x) ≤ h(x) for all x in some open interval containing a, except possibly at x = a itself. Suppose that lim x→a f(x) = lim x→a h(x) = L. Then there exists lim x→a g(x) = L. This theorem says that if g(x) is squeezed between f(x) and h(x) near a, and if f and h have the same limit L at a, then g is forced to have the same limit L at a. CALCULUS I - AY2022/23 62 / 109
  63. 63. Example Evaluate the value of the limit lim x→0 x sin 1 x . This plot leads us to expect that lim x→0 x sin 1 x = 0. But this limit is not calculated by a combination of the various limit laws. Instead we need the Squeeze Theorem to determine the limit. CALCULUS I - AY2022/23 63 / 109
  64. 64. Solution. Usually when we use the Squeeze Theorem we have two questions to answer: (1) What should we set f(x) to be? (2) What should we set h(x) to be? Answers. f(x) = −|x| and h(x) = |x|. Also we denote g(x) = x sin 1 x . These functions work well for the Squeeze Theorem. CALCULUS I - AY2022/23 64 / 109
  65. 65. • Finite limits as x → ±∞ Convention. The symbol for infinity ∞ does not represent a real number. We use ∞ to describe the behavior of a function when the values in its domain or range outgrow all finite bounds. For example, the function f(x) = 1 x is defined for all x 6= 0. CALCULUS I - AY2022/23 65 / 109
  66. 66. Observations. When x 0 and becomes increasingly large, 1 x becomes increasingly small. When x 0 and its magnitude becomes increasingly large, 1 x again becomes small. We summarize these observations by saying that f(x) = 1 x has limit 0 as x → ∞ or x → −∞, or that 0 is a limit of f(x) = 1 x at (positive) infinity and at negative infinity. CALCULUS I - AY2022/23 66 / 109
  67. 67. Definitions. We say that f(x) has the limit L as x tends to infinity and write lim x→∞ f(x) = L, if for every number ε 0, there exists a corresponding number M 0, such that for all x in the domain of f x M =⇒ |f(x) − L| ε. f(x) has the limit L as x tends to negative infinity and write lim x→−∞ f(x) = L, if for every number ε 0, there exists a corresponding number N 0, such that for all x in the domain of f x N =⇒ |f(x) − L| ε. CALCULUS I - AY2022/23 67 / 109
  68. 68. Limits at infinity have properties similar to those of finite limits. Theorem All the Limit Laws (for finite limits) are true when we replace lim x→a by lim x→∞ or lim x→−∞ . That is, the variable x may approach a finite number or ±∞ Example lim x→±∞ 1 + 2π √ 5 x2 ! = lim x→±∞ 1 + lim x→±∞ 2π √ 5 · 1 x · 1 x = lim x→±∞ 1 + lim x→±∞ 2π √ 5 · lim x→±∞ 1 x · lim x→±∞ 1 x = 1 + 2π √ 5 · 0 · 0 = 1 + 0 = 1. CALCULUS I - AY2022/23 68 / 109
  69. 69. • Infinite limits Look at the following graph of a function. It is clear that for any given horizontal line y = M, we can find a number δ 0 so that if we restrict x to lie in the interval (a − δ, a + δ) but x 6= a, then the curve y = f(x) lies above the line y = M. CALCULUS I - AY2022/23 69 / 109
  70. 70. This leads us to the definition. Definition Let f(x) be a function defined on some open interval that contains a, except possibly at a itself. Then lim x→a f(x) = ∞ (or also +∞) means that for every positive number M there is a positive number δ such that 0 |x − a| δ =⇒ f(x) M. This says that the values of f(x) can be made arbitrarily large (larger than any given number M) by requiring x to be close enough to a (within a distance δ, where δ(M) depends on M, but with x 6= a). You can see that if a larger M is chosen, then a smaller δ may be required. CALCULUS I - AY2022/23 70 / 109
  71. 71. Comment. This does not mean that ∞ is a number. Nor does it mean that the limit exists. It simply expresses the particular way in which the limit does not exist: f(x) can be made as large as we like by taking x close enough to 0. CALCULUS I - AY2022/23 71 / 109
  72. 72. Similarly, we can the negative infinite limit. Definition Let f(x) be a function defined on some open interval that contains a, except possibly at a itself. Then lim x→a f(x) = −∞ means that for every negative number N there is a positive number δ such that 0 |x − a| δ =⇒ f(x) N. This says that the values of f(x) can be made arbitrarily small (smaller than any given number N) by requiring x to be close enough to a (within a distance δ, where δ(N) depends on N, but with x 6= a). CALCULUS I - AY2022/23 72 / 109
  73. 73. Note. We may read lim x→a f(x) = ±∞ as the limit of f(x), as x tends to (approaches) a, is (plus) positive or (minus) negative infinity. Exercise Prove that lim x→0 1 x2 = ∞. CALCULUS I - AY2022/23 73 / 109
  74. 74. • One-sided infinite limits Similar definitions can be given for the one-sided infinite limits lim x→a− f(x) = ∞ lim x→a+ f(x) = ∞. lim x→− a f(x) = −∞ lim x→a+ f(x) = −∞, remembering that x → a− means that we consider only values of x a, and similarly x → a+ means that we consider only x a. CALCULUS I - AY2022/23 74 / 109
  75. 75. CALCULUS I - AY2022/23 75 / 109
  76. 76. • Infinite Limits at Infinity Now we consider the last type of limits, the infinite limits at infinity. The notation lim x→∞ f(x) = ∞ means f(x) becomes large as x becomes large. Similarly we have, lim x→−∞ f(x) = ∞, lim x→∞ f(x) = −∞, lim x→−∞ f(x) = −∞. CALCULUS I - AY2022/23 76 / 109
  77. 77. We have the following definitions. Definition Let f be a function defined on some interval (a, ∞). We write lim x→∞ f(x) = ∞, if for every number M 0, there exists a number N 0, such that x N =⇒ f(x) M. Similar definitions apply when the symbol ∞ is replaced by −∞. CALCULUS I - AY2022/23 77 / 109
  78. 78. Exercise Show that lim x→∞ x3 = ∞ lim x→−∞ x3 = −∞. CALCULUS I - AY2022/23 78 / 109
  79. 79. Outline 1 Functions 2 Limits 3 Continuity CALCULUS I - AY2022/23 79 / 109
  80. 80. Summary In this section, we study the following topics: Continuity of a function at a point; continuity test Continuous functions and their properties Important classes of continuous functions The application: Intermediate Value Theorem for continuous functions CALCULUS I - AY2022/23 80 / 109
  81. 81. Observation We noticed previously that the limit of a function as x tends to a can often be found simply by calculating the value of the function at a. We already studied the “Direct Substitution Property”: If f is a polynomial or a rational function and a is in the domain of f, then lim x→a f(x) = f(a). For example, if f(x) = x + 1 x2 − 2 , then lim x→1 f(x) = lim x→1 x + 1 x2 − 2 = f(1) = −2. CALCULUS I - AY2022/23 81 / 109
  82. 82. • Continuity It is true that for many functions f(x) we can obtain the limit lim x→a f(x) just by evaluating the function at the point a we are taking the limit. These are actually very special types of functions: the continuous functions. Definition. Let f be a function and let a ∈ R. This function is said to be continuous at a, if lim x→a f(x) = f(a). right-continuous (or continuous from the right) at a, if lim x→a+ f(x) = f(a). left-continuous (or continuous from the left) at a, if lim x→a− f(x) = f(a). CALCULUS I - AY2022/23 82 / 109
  83. 83. Note. The equation in the definition is sometimes called the direct substitution property. The intuitive idea of a continuous function is that the graph can be drawn without lifting the pen off the paper. Or, a small change in x away from x = a only leads to a small change in f(x). Below is a continuity at points x = a, x = b, and x = c CALCULUS I - AY2022/23 83 / 109
  84. 84. In the figure below, the function is not continuous at points x = 1, x = 2, and x = 4. CALCULUS I - AY2022/23 84 / 109
  85. 85. Example The function f(x) = √ 4 − x2 is continuous over its domain [−2, 2]. It is right-continuous at x = −2, and left-continuous at x = 2. CALCULUS I - AY2022/23 85 / 109
  86. 86. We summary continuity at an interior point in the form of a test, which can be easily obtained from the definition. Continuity Test A function f(x) is continuous at the point x = a if the following three conditions are true: (i) f(a) is defined (that is, a is in the domain of f) (ii) The limit lim x→a f(x) exists (that is, f has a limit as x → a) (iii) lim x→a f(x) = f(a) (that is, the limit equals the function value). Note. For one-side continuity and continuity at an endpoint of an interval, the limits in parts (ii) and (iii) of the test should be replaced by the appropriate one-side limits. CALCULUS I - AY2022/23 86 / 109
  87. 87. • Discontinuity If a function f is defined near a (that is, f is defined on an open interval containing a, except perhaps at a), and f is not continuous at a, then we say that f is discontinuous at a (or f has a discontinuity at a). Comments. From the Continuity Test, it follows that a function f is NOT continuous at a, if and only if either of the following conditions fail to hold: f(a) is not defined. The limit lim x→a f(x) does not exist. lim x→a f(x)6=f(a). CALCULUS I - AY2022/23 87 / 109
  88. 88. Exercise Figure below shows the graph of a function f. At which numbers is f discontinuous? Why? CALCULUS I - AY2022/23 88 / 109
  89. 89. • Classification of discontinuity Suppose a function f(x) is defined near a, but it is discontinuous at x = a. If the limit lim x→a f(x) = L, but L 6= f(a), then a is called a removable discontinuity. This discontinuity can be removed to make f continuous at x = a, or more precisely, the function g(x) =    f(x) x 6= a L x = a is continuous at x = a. CALCULUS I - AY2022/23 89 / 109
  90. 90. CALCULUS I - AY2022/23 90 / 109
  91. 91. If the two one-sided limits exist and are finite, but are not equal (and hence the limit L does not exist), then a is called a jump discontinuity (or discontinuity of the first kind). For this type of discontinuity, the function f may have any value at a. CALCULUS I - AY2022/23 91 / 109
  92. 92. If one of the two one-sided limits does not exist or is infinite, then a is called an essential discontinuity (or discontinuity of the second kind). CALCULUS I - AY2022/23 92 / 109
  93. 93. Exercise Classify discontinuity of the following functions at x0 = 1. (a) f(x) =          x2, x 1 0, x = 1 2 − x, x 1 (b) f(x) =          x2, x 1 0, x = 1 2 − (1 − x)2, x 1 (c) f(x) =            sin 5 x − 1 , x 1 0, x = 1 1 x − 1 , x 1 The graphs of these functions are shown in the next slide. CALCULUS I - AY2022/23 93 / 109
  94. 94. CALCULUS I - AY2022/23 94 / 109
  95. 95. • Properties of continuous functions Using Limit Laws, we can prove that algebraic combinations of continuous functions are continuous everywhere they are defined. Theorem If f, g are continuous at x = a, then so are the following: 1 Sums, differences: f ± g 2 Constant multiple: k · f, for any number k 3 Products: f · g 4 Quotients: f g , provided that g(a) 6= 0 5 Powers: fn, where n is a positive integer 6 Roots: n √ f, provided that it is defined on an open interval containing a, where n is a positive integer. CALCULUS I - AY2022/23 95 / 109
  96. 96. • Continuity on an interval So far we have only been discussing continuity at a single point, while in general we want to describe the continuity behavior of a function throughout its entire domain Definition A function f is continuous on an interval if it is continuous at every number in the interval. If f is defined only on one side of an endpoint of the interval, we understand continuous at the endpoint to mean continuous from the right or continuous from the left. Note that by interval we are including all of the following possibilities: (a, b), (a, b], [a, b), [a, b], (−∞, b), (−∞, b], (a, ∞), [a, ∞), (−∞, ∞). CALCULUS I - AY2022/23 96 / 109
  97. 97. Example The function f(x) = 2 − √ 1 − x2 is continuous on the interval [−1, 1]. Solution. - If −1 a 1, then using the Limit Laws, we have lim x→a f(x) = lim x→a (2 − p 1 − x2) = 2 − lim x→a p 1 − x2 = 2 − q lim x→a (1 − x2) = 2 − p 1 − a2 = f(a). Thus f is continuous at a. - At the end points ±1, similar calculations show that lim x→−1+ f(x) = 2 = f(−1) and lim x→1− f(x) = 2 = f(1). So f is continuous from the right at −1 as well as from the left at 1. So conclusion: f is continuous on [−1, 1]. CALCULUS I - AY2022/23 97 / 109
  98. 98. • Continuity of important functions It turns out that most of the familiar functions are continuous at every number in their domains. Polynomials: P(x) = n X i=0 aixi Power functions: f(x) = xa Rational functions: f(x) = P(x) Q(x) Algebraic functions Trigonometric functions: cos x, sin x, tan x, cot x, sec x, csc x Exponential functions: f(x) = ax (a 0, 6= 1) Logarithmic functions: f(x) = loga x (a 0, 6= 1) CALCULUS I - AY2022/23 98 / 109
  99. 99. We have the following principle. The general principle. Any function constructed from continuous functions by doing additions, multiplications, and divisions, is itself going to be continuous at every point of its domain. This principle shows how to build up complicated continuous functions from simple ones by doing additions, multiplications, and divisions. CALCULUS I - AY2022/23 99 / 109
  100. 100. Another way of combining continuous functions f and g to get a new continuous function is to form the composite function f ◦ g. This fact is a consequence of the following result. Theorem If f is continuous at b and lim x→a g(x) = b, then lim x→a f g(x) = f(b). In other words, lim x→a f g(x) = f lim x→a g(x) . This theorem says that a limit symbol lim x→a can be moved through a function symbol f if the function is continuous and the limit exists. In other words, the order of these two symbols can be reversed. CALCULUS I - AY2022/23 100 / 109
  101. 101. • Continuity of composite functions Applying the previous theorem, we can get the fact that a continuous function f of a continuous function g is a continuous function f ◦ g. More precisely, we have the following result about composition of continuous functions. Theorem If g is continuous at a and f is continuous at g(a), then the composite function f ◦ g given by (f ◦ g)(x) = f g(x) is continuous at a. CALCULUS I - AY2022/23 101 / 109
  102. 102. Example Where is the following function continuous? F(x) = 1 √ x2 + 7 − 4 . Solution. Note that F can be broken up as the composition of four continuous functions: F = f ◦ g ◦ h ◦ k or F(x) = f g h(k(x)) , where f(x) = 1 x , g(x) = x − 4, h(x) = √ x, k(x) = x2 + 7. Since each of these functions is continuous on its domain, F is continuous on its domain, which is n x ∈ R : p x2 + 7 6= 4 o = {x ∈ R : x 6= ±3} = (−∞, −3)∪(−3, 3)∪(3, ∞) CALCULUS I - AY2022/23 102 / 109
  103. 103. • Application of continuity Let us consider a function f(x) with domain [a, b] whose graph is shown below. I have concealed part of the graph under a box. Now I tell you that the graph of f(x) intersect the line y = N at no point. Question. What can you conclude? Answer. f(x) cannot be continuous on the whole interval [a, b]. CALCULUS I - AY2022/23 103 / 109
  104. 104. • Intermediate Value Theorem The observation we just made is actually an important theorem about an important property of continuous functions. The Intermediate Value Theorem for a continuous function Let f be a function, and a, b, N real numbers with a b. Suppose f is continuous on the closed interval [a, b] f(a) 6= f(b) N is in between f(a) and f(b). Then there exists a number c ∈ (a, b) such that f(c) = N. CALCULUS I - AY2022/23 104 / 109
  105. 105. The IVT for continuous functions states that a continuous function takes on every intermediate value between the function values f(a) and f(b). Note that the value N can be taken on once (as in part (a)) or more than once (as in part (b)). CALCULUS I - AY2022/23 105 / 109
  106. 106. If we think of a continuous function as a function whose graph has no hole or break, then it is easy to believe that the IVT is true. In geometric terms it says that if any horizontal line y = N is given between y = f(a) and y = f(b), then the graph of f cannot jump over the line. It must intersect y = N somewhere. CALCULUS I - AY2022/23 106 / 109
  107. 107. Example Show that the equation 4x3 − 6x2 + 3x − 2 = 0 has a root in between 1 and 2. Solution. Let f(x) = 4x3 − 6x2 + 3x − 2. We have to show that there exists a number c ∈ (1, 2) such that f(c) = 0. Taking a = 1, b = 2, and N = 0 in the IVT, we have f(1) = 4 − 6 + 3 − 2 = −1 0 f(2) = 32 − 24 + 6 − 2 = 12 0. Thus f(1) 0 f(2); that is, N = 0 is a number between f(1) and f(2). CALCULUS I - AY2022/23 107 / 109
  108. 108. Now f is continuous since it is a polynomial, so the IVT says that there is a number c ∈ (1, 2) such that f(c) = 0. In other words, the equation 4x3 − 6x2 + 3x − 2 = 0 has at least one root c in the interval (1, 2). Comment. In fact, we can locate a root more precisely by using the IVT again. For instance, it is easy to check that f(1.2) = −0.128 0 and f(1.3) = 0.548 0, and so a root must be in between 1.2 and 1.3. CALCULUS I - AY2022/23 108 / 109
  109. 109. END OF CHAPTER 1 CALCULUS I - AY2022/23 109 / 109

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